You have found the following ages (in years) of all 4 snakes at your local zoo: $ 1,\enspace 26,\enspace 8,\enspace 2$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{1 + 26 + 8 + 2}{{4}} = {9.3\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $1$ year $-8.3$ years $68.89$ years $^2$ $26$ years $16.7$ years $278.89$ years $^2$ $8$ years $-1.3$ years $1.69$ years $^2$ $2$ years $-7.3$ years $53.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{68.89} + {278.89} + {1.69} + {53.29}} {{4}} $ $ {\sigma^2} = \dfrac{{402.76}}{{4}} = {100.69\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{100.69\text{ years}^2}} = {10\text{ years}} $ The average snake at the zoo is 9.3 years old. There is a standard deviation of 10 years.